Reading Difficulty: 4.5/5

Prerequisites: Basic calculus on curves and basic probability

When dealing with the perimeter or surface area (or generally, boundary measure) of a convex set, you can instead examine its expected length/area/etc. of its projection unto a subspace (i.e. a random line, a random plane, etc.).

First, allow me to convince you that there is indeed such a nice relation in the 2D case. Let us use for perimeter.

THEOREM: Let be a convex set and let be a random unit vector, uniformly distributed. If is the line through and , then:
For some universal constant .

Proof. Unsurprisingly, this is an application of Tonelli. Let be a parametrization of . Then recall that:
So we somehow want to get there.

Note that the point stays on line as varies from to , and in doing so traces out an interval on , visiting each point on that interval twice (because it has to go back and forth). This is technically a curve, and so if we associate with the real line then we can find the "length" of this curve to be:
Note that the is there because, again, we're double-counting the interval length by finding the length of the "curve traced out by the projection".

Anyways this is pretty nice because we can now mess with the right side:
Now we take the expectation:
And hey, expectations are just integrals, and the integrand is non-negative, so we can swap using Tonelli:
This expectation is actually nice. To argue that, we switch gears to geometry. This expectation is really just the expected distance from the origin of the projection of unto the line . We can instead view this as taking a random point on a circle centered at and radius and computing the expected absolute value of its -coordinate. I don't actually need to compute this to prove the theorem, but I'll do it anyway. By four-fold symmetry this is just . So our nasty integral is really just:
Whoa, magic! Thus the theorem has been proven, and we have shown that if you care.

Sanity Check: Verify that this is theorem is true when is just the unit disk. Indeed

Cool. Now let's obliterate some hard math problems.

Problem 1: Recall the famous math puzzle which goes something like, "you have a rope wrapped tightly around a basketball and a rope wrapped tightly around Earth. You increase the length of each rope so that it hovers 1 foot above the surface. Which rope's length increased more?" If you do the math, the answer is that they increase equally, each by .

Prove that this remains true for convex sets instead of circles. That is, for convex, we have that where .

Solution

For any line , the projection of unto will always be exactly 2 greater than the projection of unto . So:
Oops.

Problem 2: are convex with . Prove that .

Proof. Try it yourself first

Problem 3: We can, in fact, generalize Problem 1. Recall the for two sets , their "Minkowski Sum" is given by:
Show that perimeter is additive on convex sets (!!!). That is, whenever are convex.

Solution: Because of how Minkowski sum is defined, the sum of the lengths of the projections of unto some line is just equal to the length of the projection of unto line . It is then trivial.

Problem 4: To mail a box, a delivery service charges you the sum of the box's dimensions. Show that you can't cheat by putting a box in a box.

(That is, if Box 1 is inside Box 2, then the sum of dimensions of Box 1 is at most the sum of dimensions of Box 2.)

Solution: The key idea is to notice that if form three perpendicular edges of some box, then the sum of their "heights" (their projections unto the -axis) is equal to the "height" of the box (the box's projection unto the -axis). Similarly, the sum of the lengths of the projections of unto an arbitrary line must be precisely the length of the projection of the whole box unto .

Using ideas from the theorem, you can show that the expected length of the projection of a segment (that's in space) unto a randomly chosen line (which is randomly chosen in space by, say, unif. choosing a point on the surface of the unit ball) is a universal constant times the length of the segment.

Now we're done because if Box 1 is inside Box 2, then on any random line , the projection of Box 1 unto line lies inside the projection of Box 2 unto line , so by the assertions in the previous paragraphs, we must conclude that, since the length of Box 1's projection is always at most the length of Box 2's projection, it follows that the sum of dimensions of Box 1 is at most the sum of dimensions of Box 2.


Problem 5 (Barbier's Theorem): Every convex shape of constant width must have perimeter .

Proof. I mean... duh.